Chapter 11, Problem 11.44QA

Interpretation Introduction

To find

The concentration of given elements in molality units.

Answer to Problem 11.44QA

Solution:

i. Molality of Al³⁺ is $1.9 \times {10}^{-6} m$

ii. Molality of Fe³⁺ is $7.2 \times {10}^{-7} m$

iii. Molality of Ca²⁺ is $3.3 \times {10}^{-4} m$

iv. Molality of Na⁺ is $2.3 \times {10}^{-4} m$

v. Molality of K⁺ is $3.3 \times {10}^{-5} m$

vi. Molality of Mg²⁺ is $1.4 \times {10}^{-4} m$

Explanation of Solution

1) Concept:

Here we assume $1$ kg of river water. With the mass of river water and given concentrations we will find out the mass of each element. From the mass of element and mass of river water we can find out mass of solvent. From the mass and molar mass of element we will find out moles of each element. Finally using moles and mass of solvent in kg we will find out molality of each element.

2) Formula:

$Molality= \frac{mol of solute}{mass of solvent in kg}$

A mole is the SI unit of amount chemical substance. When writing units, it is written as “mol”.

3) Given:

i. Concentration of Al³⁺ = $0.050 mg/kg$

ii. Concentration of Fe³⁺ = $0.040 mg/kg$

iii. Concentration of Ca²⁺ = $13.4 mg/kg$

iv. Concentration of Na⁺ = $5.2 mg/kg$

v. Concentration of K⁺ = $1.3 mg/kg$

vi. Concentration of Mg²⁺ = $3.4 mg/kg$

Conversion factor:

$1 kg=1000 g$

$1 g=1000 mg$

4) Calculations:

We will convert mass of river water from kilogram to gram.

$1 kg \times \frac{1000 g}{1 kg}=1000 g$

Calculation of molality of Al³⁺.

We calculate the mass of Al³⁺ using $1$ kg mass of river water and given concentration of Al³⁺.

$Mass of {Al}^{3+}=1 kg \times \frac{0.050 mg}{1 kg}=0.050 mg$

We will convert this mass from mg to kg using the conversion factor between them.

$0.050 mg \times \frac{1 g}{1000 mg}=5.0 \times {10}^{-5} g$

From mass of river water and mass of element we will calculate mass of solvent.

$Mass of solvent=Mass of solution-Mass of solute$

$Mass of solvent=1000 g- 5.0 \times {10}^{-5} g=999.99995 g$

Convert mass of solvent from gram to kilogram using the conversion factor between them.

$999.99995 g \times \frac{1 kg}{1000 g}=0.99999995 kg$

Now we calculate moles of solute using $5.0 \times {10}^{-5} g$ and molar mass of Al³⁺ ($26.982 g/mol$)

$5.0 \times {10}^{-5} g \times \frac{1 mol}{26.982 g}=1.853 \times {10}^{-6} mol$

Finally using mole solute and mass of solvent we can calculate molality of Al³⁺.

$m= \frac{1.853 \times {10}^{-6} mol}{0.99999995 kg}= 1.853 \times {10}^{-6}\frac{mol}{kg}of {Al}^{3+}$

The molality of Al³⁺ is $1.9 \times {10}^{-6} m$.

Calculation of molality of Fe³⁺.

We calculate the mass of Fe³⁺ using 1 kg mass of river water and given concentration of Fe³⁺.

$Mass of {Fe}^{3+}=1 kg \times \frac{0.040 mg}{1 kg}=0.040 mg$

We will convert this mass from mg to kg using the conversion factor between them.

$0.040 mg \times \frac{1 g}{1000 mg}=4.0 \times {10}^{-5} g$

From mass of river water and mass of element we will calculate mass of solvent.

$Mass of solvent=Mass of solution-Mass of solute$

$Mass of solvent=1000 g- 4.0 \times {10}^{-5} g=999.99996 g$

Convert mass of solvent from gram to kilogram using the conversion factor between them.

$999.99996 g \times \frac{1 kg}{1000 g}=0.99999996 kg$

Now we calculate moles of solute using $4.0 \times {10}^{-5} g$ and molar mass of Fe³⁺ ($55.845 g/mol$)

$4.0 \times {10}^{-5} g \times \frac{1 mol}{55.845 g}=7.163 \times {10}^{-7} mol$

Finally using mole solute and mass of solvent we can calculate molality of Fe³⁺.

$m= \frac{7.163 \times {10}^{-7} mol }{0.99999996 kg}= 7.163 \times {10}^{-7} \frac{mol}{kg}of {Fe}^{3+}$

The molality of Fe³⁺ is $7.2 \times {10}^{-7} m$.

Calculation of molality of Ca²⁺.

We calculate the mass of Ca²⁺ using 1 kg mass of river water and given concentration of Ca²⁺.

$Mass of {Ca}^{2+}=1 kg \times \frac{13.4 mg}{1 kg}=13.4 mg$

We will convert this mass from mg to kg using the conversion factor between them.

$13.4 mg \times \frac{1 g}{1000 mg}=1.34 \times {10}^{-4} g$

From mass of river water and mass of element we will calculate mass of solvent.

$Mass of solvent=Mass of solution-Mass of solute$

$Mass of solvent=1000 g- 1.34 \times {10}^{-4} g=999.9866 g$

Convert mass of solvent from gram to kilogram using the conversion factor between them.

$999.9866 g \times \frac{1 kg}{1000 g}=0.9999866 kg$

Now we calculate moles of solute using $1.34 \times {10}^{-4} g$ and molar mass of Ca²⁺ ($40.078 g/mol$)

$1.34 \times {10}^{-4} g \times \frac{1 mol}{40.078 g}=3.34 \times {10}^{-4} mol$

Finally using mole solute and mass of solvent we can calculate molality of Ca²⁺.

$m= \frac{3.34 \times {10}^{-4} mol }{0.9999866 kg}= 3.34 \times {10}^{-4} \frac{mol}{kg}of {Ca}^{2+}$

The molality of Ca²⁺ is $3.3 \times {10}^{-4} m$

Calculation of molality of Na⁺.

We calculate the mass of Na⁺ using 1 kg mass of river water and given concentration of Na⁺.

$Mass of {Na}^{+}=1 kg \times \frac{5.2 mg}{1 kg}=5.2 mg$

We will convert this mass from mg to kg using the conversion factor between them.

$5.2 mg \times \frac{1 g}{1000 mg}=5.2 \times {10}^{-3} g$

From mass of river water and mass of element we will calculate mass of solvent.

$Mass of solvent=Mass of solution-Mass of solute$

$Mass of solvent=1000 g- 5.2 \times {10}^{-3} g=999.9948 g$

Convert mass of solvent from gram to kilogram using the conversion factor between them.

$999.9948 g \times \frac{1 kg}{1000 g}=0.9999948 kg$

Now we calculate moles of solute using $1.34 \times {10}^{-4} g$ and molar mass of Na⁺ ($22.99 g/mol$)

$5.2 \times {10}^{-3} g \times \frac{1 mol}{22.99 g}=2.26 \times {10}^{-4} mol$

Finally using mole solute and mass of solvent we can calculate molality of Na⁺.

$m= \frac{2.26 \times {10}^{-4} mol }{0.9999948 kg}= 2.26\times {10}^{-4} \frac{mol}{kg}of {Na}^{+}$

The molality of Na⁺ is $2.3\times {10}^{-4} m$.

Calculation of molality of K⁺.

We calculate the mass of K⁺ using 1 kg mass of river water and given concentration of K⁺.

$Mass of K^{+}=1 kg \times \frac{1.3 mg}{1 kg}=1.3 mg$

We will convert this mass from mg to kg using the conversion factor between them.

$1.3 mg \times \frac{1 g}{1000 mg}=1.3 \times {10}^{-3} g$

From mass of river water and mass of element we will calculate mass of solvent.

$Mass of solvent=Mass of solution-Mass of solute$

$Mass of solvent=1000 g- 1.3 \times {10}^{-3} g=999.9987 g$

Convert mass of solvent from gram to kilogram using the conversion factor between them.

$999.9987 g \times \frac{1 kg}{1000 g}=0.9999987 kg$

Now we calculate moles of solute using $1.3 \times {10}^{-3} g$ and molar mass of K⁺ ($39.098 g/mol$)

$1.3 \times {10}^{-3} g \times \frac{1 mol}{39.098 g}=3.32 \times {10}^{-5} mol$

Finally using mole solute and mass of solvent we can calculate molality of K⁺.

$m= \frac{3.32 \times {10}^{-5} mol }{0.9999987 kg}= 3.32 \times {10}^{-5} \frac{mol}{kg}of K^{+}$

The molality of K⁺ is $3.3 \times {10}^{-5} m$.

Calculation of molality of Mg²⁺.

We calculate the mass of Mg²⁺ using 1 kg mass of river water and given concentration of Mg²⁺.

$Mass of {Mg}^{2+}=1 kg \times \frac{3.4 mg}{1 kg}=3.4 mg$

We will convert this mass from mg to kg using the conversion factor between them.

$3.4 mg \times \frac{1 g}{1000 mg}=3.4 \times {10}^{-3} g$

From mass of river water and mass of element we will calculate mass of solvent.

$Mass of solvent=Mass of solution-Mass of solute$

$Mass of solvent=1000 g- 3.4 \times {10}^{-3} g=999.9966 g$

Convert mass of solvent from gram to kilogram using the conversion factor between them.

$999.9966 g \times \frac{1 kg}{1000 g}=0.9999966 kg$

Now we calculate moles of solute using $3.4 \times {10}^{-3} g$ and molar mass of Mg²⁺ ($24.305 g/mol$)

$3.4 \times {10}^{-3} g \times \frac{1 mol}{24.305 g}=1.4 \times {10}^{-4} mol$

Finally using mole solute and mass of solvent we can calculate molality of Mg²⁺.

$m= \frac{1.4 \times {10}^{-4} mol }{0.9999966 kg}= 1.4 \times {10}^{-4} \frac{mol}{kg}of {Mg}^{2+}$

The molality of Mg²⁺ is $1.4 \times {10}^{-4} m$

Conclusion

With the mass of river water and given concentrations of elements we find out the mass of each element. From the mass of element and mass of river water we find out mass of solvent. From the mass and molar mass of element we find out moles of each element. Finally using moles and mass of solvent in kg we find out molality of each element.